This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#include "seats.h"
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
#define all(v) v.begin(), v.end()
typedef long long ll;
const int b = 1 << 20;
int n, lazy[b * 2], seg[b * 2], cnt[b * 2];
vector<int> Y, X;
vector<vector<int>> v;
void upd_lazy(int ix, int l, int r){
if(l < n && seg[ix] == b) seg[ix] = 0, cnt[ix] = min(n - 1, r) - l + 1;
if(!lazy[ix]) return;
seg[ix] += lazy[ix];
if(l ^ r){
lazy[ix * 2] += lazy[ix];
lazy[ix * 2 + 1] += lazy[ix];
}
lazy[ix] = 0;
return;
}
void upd(int ix, int nl, int nr, int l, int r, int v){
upd_lazy(ix, nl, nr);
if(nl > r || nr < l) return;
if(nl >= l && nr <= r){
lazy[ix] += v;
upd_lazy(ix, nl, nr);
return;
}
int m = (nl + nr) / 2;
upd(ix * 2, nl, m, l, r, v);
upd(ix * 2 + 1, m + 1, nr, l, r, v);
seg[ix] = min(seg[ix * 2], seg[ix * 2 + 1]);
cnt[ix] = (seg[ix * 2] == seg[ix] ? cnt[ix * 2] : 0) + (seg[ix * 2 + 1] == seg[ix] ? cnt[ix * 2 + 1] : 0);
return;
}
void calc(int y, int x, int p){
vector<int> t;
for(int i = -1; i <= 0; i++)
for(int j = -1; j <= 0; j++)
t.emplace_back(v[y + i][x + j]);
sort(all(t));
upd(1, 0, b - 1, t[0], t[1] - 1, p);
upd(1, 0, b - 1, t[2], t[3] - 1, p);
return;
}
void give_initial_chart(int H, int W, vector<int> R, vector<int> C) {
Y = R; X = C;
n = H * W;
v = vector<vector<int>> (H + 2, vector<int> (W + 2, n));
for(int i = 0; i < n; i++) v[++Y[i]][++X[i]] = i;
for(int i = 1; i < b * 2; i++) seg[i] = b;
for(int i = 1; i <= H + 1; i++)
for(int j = 1; j <= W + 1; j++) calc(i, j, 1);
return;
}
int swap_seats(int i1, int i2) {
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++){
calc(Y[i1] + i, X[i1] + j, -1);
calc(Y[i2] + i, X[i2] + j, -1);
}
swap(Y[i1], Y[i2]);
swap(X[i1], X[i2]);
swap(v[Y[i1]][X[i1]], v[Y[i2]][X[i2]]);
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++){
calc(Y[i1] + i, X[i1] + j, 1);
calc(Y[i2] + i, X[i2] + j, 1);
}
if(seg[1] == 4) return cnt[1];
return 0;
}
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